Question: You have found the following ages (in years) of 6 bears. The bears are randomly selected from the 50 bears at your local zoo: $ 14,\enspace 18,\enspace 11,\enspace 22,\enspace 2,\enspace 11$ Based on your sample, what is the average age of the bears? What is the variance? You may round your answers to the nearest tenth.
Because we only have data for a small sample of the 50 bears, we are only able to estimate the population mean and variance by finding the sample mean $({\overline{x}})$ and sample variance $({s^2})$ To find the sample mean , add up the values of all $6$ samples and divide by $6$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6$ To compensate for this underestimation, rather than simply averaging the squared deviations from the mean , we total them and divide by $n - 1$ $ {s^2} = \dfrac{\sum\limits_{i=1}^{{n}} (x_i - {\overline{x}})^2}{{n - 1}} $ $ {s^2} = \dfrac{{1} + {25} + {4} + {81} + {121} + {4}} {{6 - 1}} $ $ {s^2} = \dfrac{{236}}{{5}} = {47.2\text{ years}^2} $ We can estimate that the average bear at the zoo is 13 years old. There is a variance of 47.2 years $^2$.